Question

12.Prove that : (sinA + secA)² + (cosA + cosecA)²= (1+ secA.cosecA)²​

Answer 1

Step-by-step explanation:

your answer in the attached file

Answer 2

Question  :

12.Prove that : (sinA + secA)² + (cosA + cosecA)²= (1+ secA.cosecA)²​

Solution :

$\longmapsto\sf (sinA + secA)^2 + (cosA + cosecA)^2 \\\\ \\ \longmapsto\sf sin^2A + 2sinA.secA + sec^2A + cos^2A + 2cosA.cosecA + cosec^2A \qquad[\because \ (a + b)^2 = a^2 + 2ab + b^2] \\\\ \\ \longmapsto\sf (sin^2A + cos^2A) + (sec^2A + cosec^2A) + 2 \bigg[\dfrac{sinA}{cosA} + \dfrac{cosA}{sinA} \bigg] \\\\ \\ \longmapsto\sf 1 + \bigg[\dfrac{1}{cos^2A} + \dfrac{1}{sin^2A} \bigg] + 2 \bigg[\dfrac{sin^2A + cos^2A}{sinAcosA} \bigg] \qquad[\because sin^2A + cos^2A = 1]$

$\longmapsto\sf 1 + \bigg[\dfrac{sin^2A + cos^2A}{sin^2Acos^2A} \bigg] + 2 \bigg[\dfrac{1}{sinA} \times \dfrac{1}{cosA} \bigg] \\\\ \\ \longmapsto\sf 1 + \dfrac{1}{sin^2A} \times \dfrac{1}{cos^2A} + 2. \ cosecA . \ secA \\\\ \\ \longmapsto\sf 1 + cosec^2A.sec^2A + 2. \ cosecA. \ secA \qquad\bigg[\because \dfrac{1}{sin^2A} = cosec^2A ; \dfrac{1}{cos^2A} = sec^2A \bigg] \\\\ \\ \longmapsto\boxed{\boxed{\bold{(1 + secA. cosecA)^2 }}}$

${\: \: \: \: \: \: \: } \therefore\underline{\boxed{\bold{L.H.S \ = \ R.H.S \qquad[Hence \ proved] }}}$